R-UFD, F-fraction field, f monic in F[X], f(a)=0=> a in R

[UD1]

Homework Statement

Let R be a UFD and let F be the field of quotients/fractions for R. If f(a)=0, where f is in R[X] is monic and a is in F, show that a is in R.

The Attempt at a Solution

On one hand if a is F, we can write a=m/n for some m and n in R, where in fact m and n are comprime. Then

f(a)=(m/n)^k+r_{n-1}(m/n)^{k-1}+...+r_1(m/n)+r_0.

Another idea is to use Gauss's lemma. Since f has a root, there exist some g and h from F[X] such that f=gh. But then f=g_0 * h_0 for some polynomials g_0, h_0 from R[X]. Then for one of them, say g_0, we have g_0(a)=0. However that doesn't mean that a is in R.

 

[Tedjn]

Your first idea seems the right step to take. Since the expression f(a) is zero, you can clear denominators and turn the problem into a divisibility problem in R.